 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-1/12f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Since

f (0) = 0

\frac{f ( h ) - f ( 0 )}{h} = h sign (h) cos \frac{π}{h} = ∣ h ∣ cos \frac{π}{h} (h \neq = 0) .

1

, so this quotient lies between

0

and

∣ h ∣

f is differentiable at 0, f^{'} (0) = 0.

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